Subject: RE: [xsl] Getting node w/ lowest attribute value From: "Bordeman, Chris" <Chris.Bordeman@xxxxxxxxxxxxxxxxx> Date: Sat, 16 Aug 2008 12:42:48 -0500 |
Cool, thanks Mukul, now how do I get the 2nd lowest? -----Original Message----- From: Mukul Gandhi [mailto:gandhi.mukul@xxxxxxxxx] Sent: Friday, August 15, 2008 1:10 PM To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Subject: Re: [xsl] Getting node w/ lowest attribute value Please try this, addresses/address[(@IsActive = 'true') and (@NumOrder = min(../address[@IsActive = 'true']/@NumOrder))][1] This uses the XPath 2.0 function, 'min'. On Fri, Aug 15, 2008 at 11:22 PM, Bordeman, Chris <Chris.Bordeman@xxxxxxxxxxxxxxxxx> wrote: > Hi all. > > I have some nodes like: > > <addresses> > <address IsActive="false" NumOrder=1>[...]</address> > <address IsActive="true" NumOrder=3>[...]</address> > <address IsActive="true" NumOrder=2>[...]</address> </addresses> > > How do I get the first address node where IsActive=true AND has the > lowest value for the NumOrder attribute? > > In the above case I'd want the 3rd address node (IsActive="true" and > NumOrder=2). > > Any assistance would be appreciated. Thanks.
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